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From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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08 Jul 2012, 16:29
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From the consecutive integers 10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers? A. (10)^20 B. (10)^10 C. 0 D. –(10)^19 E. –(10)^20
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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08 Jul 2012, 16:43
catty2004 wrote: From the consecutive integers 10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A. (10)^20 B. (10)^10 C. 0 D. –(10)^19 E. –(10)^20 Select 10 odd number of times and 10 the remaining number of times, for example select 10 once and 10 nineteen times, then the product will be \(10^1*(10)^{19}=10^{20}\). Answer: E. P.S. Please read and follow: rulesforpostingpleasereadthisbeforeposting133935.html
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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19 Dec 2012, 00:32
The minimum value is the biggest whole number with a negative sign.
\((10)*(10)^{19}\)
But this is not anywhere in the answer choice, let's manipulate: \((10)(10)^{20}=(1)(10)(10)^{19}=10^{20}\)
Answer: E




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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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08 Jul 2012, 16:53
Bunuel wrote: catty2004 wrote: From the consecutive integers 10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A. (10)^20 B. (10)^10 C. 0 D. –(10)^19 E. –(10)^20 Select 10 odd number of times and 10 the remaining number of times, for example select 10 once and 10 nineteen times, then the product will be \(10^1*(10)^{19}=10^{20}\) . Answer: E. P.S. Please read and follow: rulesforpostingpleasereadthisbeforeposting133935.htmlI'm not quiet sure i understand, why choose only 10 and 10, there are other integers included in the set of 10 to 10. Also, isn't 10^1*10^(19) = 10^(18)?



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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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08 Jul 2012, 16:57
catty2004 wrote: Bunuel wrote: catty2004 wrote: From the consecutive integers 10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A. (10)^20 B. (10)^10 C. 0 D. –(10)^19 E. –(10)^20 Select 10 odd number of times and 10 the remaining number of times, for example select 10 once and 10 nineteen times, then the product will be \(10^1*(10)^{19}=10^{20}\) . Answer: E. P.S. Please read and follow: rulesforpostingpleasereadthisbeforeposting133935.htmlI'm not quiet sure i understand, why choose only 10 and 10, there are other integers included in the set of 10 to 10. Also, isn't 10^1*10^(19) = 10^(18)? We choose 10 and 10 because this way we get the least product (notice that 10^(20) is the smallest number among answer choices). Also it's \(10^1*(10)^{19}=10^{20}\) not \(10^1*(10)^{19}=10^{20}\).
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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08 Jul 2012, 17:35
Thanks for pointing that out!!! Silly me. But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_<



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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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Updated on: 09 Jul 2012, 02:00
catty2004 wrote: From the consecutive integers 10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A. (10)^20 B. (10)^10 C. 0 D. –(10)^19 E. –(10)^20 The consecutive integers from 10 to 10 are: 10, 9, 8, 7 ... 1, 0, 1, ... 9, 10 You have to select 20 integers with repetition. So you can select all 10. The product will be \(10^{20}\). or You can select all 10. The product will be \((10)^{20}\) (which is essentially same as \(10^{20}\)) or You can select ten 1s and ten 10s. The product will be \((10)^{10}\) or You can select 0 and any other 19 numbers. The product will be 0. or You can select 1 and nineteen 10s. The product will be \((10)^{19}\). But how will you get \( (10)^{20}\)? Editing here itself  I missed out the case in which you can get \( (10)^{20}\) (by multiplying odd number of 10 with rest of the 10s). Possibly because I misread the question. 'Least possible value of the product' to me implied the product that is not possible. Since all options a to d were possible, I overlooked that option (E) is possible too. Actually the question should read 'the smallest possible value of the product' to avoid such confusion. Since 10^20 is the smallest value that can be obtained (greatest magnitude but negative), the answer is (E).
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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09 Jul 2012, 00:38
VeritasPrepKarishma wrote: catty2004 wrote: From the consecutive integers 10 to 10 inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A. (10)^20 B. (10)^10 C. 0 D. –(10)^19 E. –(10)^20 The consecutive integers from 10 to 10 are: 10, 9, 8, 7 ... 1, 0, 1, ... 9, 10 You have to select 20 integers with repetition. So you can select all 10. The product will be \(10^{20}\). or You can select all 10. The product will be \((10)^{20}\) (which is essentially same as \(10^{20}\)) or You can select ten 1s and ten 10s. The product will be \((10)^{10}\) or You can select 0 and any other 19 numbers. The product will be 0. or You can select 1 and nineteen 10s. The product will be \((10)^{19}\). But how will you get \( (10)^{20}\)? You need to select 1 and twenty 10s but you cannot select 21 numbers. You cannot have the product as negative \(10^{20}\). Hence (E) is not possible. E is possible if you select 10 odd number of times and 10 the remaining number of times. For example: \(10^1*(10)^{19}=10^{20}\) or \(10^3*(10)^{17}=10^{20}\) or \(10^5*(10)^{15}=10^{20}\)... or \(10^{19}*(10)^{1}=10^{20}\). So, the answer is E.
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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09 Jul 2012, 01:54
catty2004 wrote: But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_< Actually, the question means "smallest possible value of the product" \(10^{20}\) is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading.
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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09 Jul 2012, 01:58
VeritasPrepKarishma wrote: catty2004 wrote: But i'm still not sure, why the other answer choices are eliminated and why E is the least possible answer?? >_< Actually, the question means "smallest possible value of the product" \(10^{20}\) is the smallest value of the given 5 options. Since it is possible, so it is the answer. I guess you were thinking 'least possible' (as was I). The wording is definitely misleading. Yes, the question means "what is the smallest possible value of the product of the 20 integers?"
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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24 Jul 2012, 20:59
i missed out the bracket for 10^20. I was assuming that value to be positive as a negative number raised to positive power is positive. Nice problem![/edit]



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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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11 Nov 2012, 18:55
Bigred2008 wrote: From the consecutive integers 10 to 10, inclusive 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of 20 integers.
A. (10)^20 B. (10)^10 C. 0 D. (10)^19 E. (10)^20 The smallest possible value of the product will be negative (since it has to be smallest) with highest absolute value i.e. the number should be as left on the number line as possible. Out of all the options, (E) is the smallest. Let's see if it is possible. If you select all 10s and only one 10, you will get \(10*(10)^{19}\) which is the same as \((10)^{20}\). So it is possible to get the product given in (E). Answer must be (E) Notice that options (A) and (B) are positive.
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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25 Aug 2014, 10:05
This question shows how important it is to carefully look at the symbols. One might be very easily deceived in thanking the question is actually saying > (10)^20.



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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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19 Nov 2014, 09:07
Hi guys,
just to make sure, picking "10" 19x and "+10" the remaining time would also work, right?
> (10)^19 * 10^1
?
Many thanks



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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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19 Nov 2014, 19:47
mott wrote: Hi guys,
just to make sure, picking "10" 19x and "+10" the remaining time would also work, right?
> (10)^19 * 10^1
?
Many thanks Yes, it does work. That is how you get \((10)^{20}\).
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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20 Jan 2015, 04:26
Hi,
I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.
So what I though it that we need an odd number of negative values to have a negative value. To have the greatest negative value we need 10 (the higest possible value), the greatest number of odd times; so in out case 19. We still have to use one number, so 10 would be the final choise, as it would make the negative 19(10) 10 times greater.
However, for me it is: 10(19) * 10 (1) instead of 10^1 * 10^19 = 10^20.
I understand the 10^1, but 10(19) is not 10^19. This is where I am lost...
So, why the powers?



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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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20 Jan 2015, 04:36
pacifist85 wrote: Hi,
I also chose the right answer quite quickly. However, I didn't use powers and I am not sure why we used powers.
So what I though it that we need an odd number of negative values to have a negative value. To have the greatest negative value we need 10 (the higest possible value), the greatest number of odd times; so in out case 19. We still have to use one number, so 10 would be the final choise, as it would make the negative 19(10) 10 times greater.
However, for me it is: 10(19) * 10 (1) instead of 10^1 * 10^19 = 10^20.
I understand the 10^1, but 10(19) is not 10^19. This is where I am lost...
So, why the powers? You are given that you need to find the product of all 20 numbers. When you multiply 10 by 10, you get \((10)^2\), not \(10*2\) (which is 10 + (10)) Just like you multiplied the last 10 you picked, you have to multiply all previous 10s too. So multiply 10, 19 times to get \(10*10*10...*10 = (10)^{19}\)
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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27 Mar 2016, 07:38
mbaspire wrote: From the consecutive integers 10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A \((10)^{20}\) B \((10)^{10}\) C 0 D \((10)^{19}\) E \((10)^{20}\) Hi, we have to choose all 10s and ensure the combination is such that the sign becomes ive.. \(10^{19}*(10)..\) or \((10)^{19}*10= 10^{20}\) E
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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27 Mar 2016, 07:38
mbaspire wrote: From the consecutive integers 10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A \((10)^{20}\) B \((10)^{10}\) C 0 D \((10)^{19}\) E \((10)^{20}\) Choose nineteen 10's and one 10 So, the product = [(10)^19][10]Notice that [(10)^19] is NEGATIVE, which means [(10)^19][10] is also NEGATIVE. So, [(10)^19][10] = [(10)^19][10]= (10)^20 = E Cheers, Brent
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Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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28 Mar 2016, 00:42
Bunuel wrote: mbaspire wrote: From the consecutive integers 10 to 10, inclusive, 20 integers are randomly chosen with repetitions allowed. What is the least possible value of the product of the 20 integers?
A \((10)^{20}\) B \((10)^{10}\) C 0 D \((10)^{19}\) E \((10)^{20}\) Hi Bunuel, I'm confused. I have referred to gmat club math book about multiply 2 number with exponents. the base should be the same to allow sum of the exponent while in this question the base is 10 & 10. So we can't sum 10^10 * (10)^19. Do I miss something? Thanks




Re: From the consecutive integers 10 to 10 inclusive, 20 integers are ran
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